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ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8

ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter12: Solutions
Section: Chapter Questions
Problem 12.107QE
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Please help using the information below calculate the limiting reagent and the theoretical yield
Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant:-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass = 138.21 gm/mol. And we make 20 m² of each. Step2 b) Solution :- Reactant-1 Potassiuma casi bonate Concentration (c) = 0·200 M = 0·200 mol/L Volume (V) = 20 mL = 20 L = 0·02 L [1L = 1000 m²] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (C) x Volume (V) = 0-200 mol x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n n = cx V = 0·220 mol x0·02 L = 0.0044 mol. Molar mass = 138.21 g/mol. NOW, mol Mass = Molex Molar mass = 0.0044 mdx138.21 m² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. √x Do 8
Transcribed Image Text:Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant:-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass = 138.21 gm/mol. And we make 20 m² of each. Step2 b) Solution :- Reactant-1 Potassiuma casi bonate Concentration (c) = 0·200 M = 0·200 mol/L Volume (V) = 20 mL = 20 L = 0·02 L [1L = 1000 m²] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (C) x Volume (V) = 0-200 mol x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n n = cx V = 0·220 mol x0·02 L = 0.0044 mol. Molar mass = 138.21 g/mol. NOW, mol Mass = Molex Molar mass = 0.0044 mdx138.21 m² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. √x Do 8
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